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The vapour pressure of water at 20 °C is 17.5 torr. What will be the no. of moles of water present in one litre of air at 20 °C and 40% relative humidity.(R =0.082 L atm/mol-k)
  • a)
    4.2 x 10-4 mole
  • b)
    4.2 x 10-6 mole
  • c)
    3.82 X 10-4 mole
  • d)
    3.82 X 10-6 mole
Correct answer is option 'C'. Can you explain this answer?
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The vapour pressure of water at 20 °C is 17.5 torr. What will be t...
Relative humidity (RH) = 
∴ Partial pressure of H2O = RH × Vapour pressure of H2O
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The vapour pressure of water at 20 °C is 17.5 torr. What will be t...
Calculation of moles of water in one litre of air at 20 °C and 40% relative humidity:

Given:
- Vapour pressure of water at 20 °C = 17.5 torr
- Total pressure of air = 1 atm
- Relative humidity = 40%
- Gas constant (R) = 0.082 L atm/mol-K

Step 1: Calculating the partial pressure of water vapor:
- The partial pressure of water vapor can be calculated using the formula:
Partial pressure of water vapor = Relative humidity * Vapour pressure of water at the given temperature
Partial pressure of water vapor = 0.4 * 17.5 torr = 7 torr

Step 2: Calculating the moles of water vapor:
- Using the ideal gas law equation:
PV = nRT
n = PV / RT
n = (7 torr * 1 L) / (0.082 L atm/mol-K * 293 K)
n = 0.3 moles
Therefore, the number of moles of water present in one litre of air at 20 °C and 40% relative humidity is 0.3 moles, which is equivalent to 3.82 x 10^-4 moles. Hence, option C is the correct answer.
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Consider two liquids B and C that form an ideal solutlion. We hold the temperature fixed at some value T that is above the freezing points of B and C .We shall plot the systems pressure P and against XB the overall mole fraction of B in the system :Where nbland nbvare the number of moles of B in the liquid and vapor phases, respectively. For a close system xB, is fixed, although nBland nBvmay vary. Let the system be enclosed in a cylinder fitted with a piston and immersed in a constant-temperature bath.To see what the P-versus-xB phase diagram looks like, let us initially set the external pressure on the piston high enough for the system to be entirely liquid (point A in figur e) As the pressure is lowered below that at A, the system eventually reaches a pressure where the liquid just begins to vaporizes (point D). At point D, the liquid has composition xlbwhere xlbat D is equal to the overall mole fraction xB since only an infinitesimal amount of liquid has vapourized. What is the composition of the first vapour that comes off ? Raoults law, Pb=xvbPobrelates the vapour-phase mole fractions to the liquid composition as follows:Where Poband Pocare the vapour pressures of pure B and pure C at T, where the systems pressure P equals the sum PB + Pc of the partial pressures, where and the vapor is assumed ideal.Let B be the more volatile component, meaning that PobPocAbove equation then shows that Xvb/XvcXlb/XlcThe vapor above an ideal solution is richer than the liquid in the more volatile component. Equations (1) and (2) apply at any pressure where liquid -vapor equilibrium exists, not just at point D. Now let us isothermally lower the pressure below point D, causing more liquid to vaporize. Eventually, WE reach point F in figure, where the last drop of liquid vaporizes. Below F, we have only vapor. For any point or the line between D and F liquid and vapor phases coexist in equilibrium.Q.Two liquids A and B have the same molecular weight and form an ideal solution. The solution has a vapor pressure of 700 Torrs at 80C. It is distilled till 2/3rd of the solution (2/3rd moles out of total moles) is collected as condensate.The composition of the condensate is xA = 0.75 and that of the residue is XA= 0.30. If the vapor pressure of the residue at 80C is 600 Torrs, find the original composition of the liquid.

Ideal Solution at Fixed TemperatureConsider two liquids B and C that form an ideal solution. We hold the temperature fixed at some value T thatis above the freezing points of B and C. We shall plot the systems pressure P against XB, the overall molefraction of B in the system :Where are the number of moles of B in the liquid and vapor phases, respectively. For a closed system XB is fixed, although may vary.Let the system be enclosed in a cylinder fitted with a piston and immersed in a constant-temperature bath. To see what the P-versus–XB phase diagram looks like, let us initially set the external pressure on the piston high enough for the system to be entirely liquid (point A in figur e) As the pressure is lowered below that at A, the system eventually reaches a pressure where the liquid just begins to vaporizes (point D). At point D, the liquid has composition at D is equal to the overall mole fraction XB since only an infinitesimal amount of liquid has vapourized. What is the composition of the first vapour that comes off ? Raoults law, relates the vapour-phase mole fractions to the liquid composition as follows :............(1)Where PB0 and PC0 are the vapour pressures of pure B and pure C at T, where the systems pressure P equals the sum PB + PC of the partial pressures, where, and the vapour is assumed ideal. ............(2)Let B be the more volatile component, meaning that . Above equation then shows that The vapour above an ideal solution is richer than the liquid in the more volatile component. Equations (1) and (2) apply at any pressure where liquid –vapour equilibrium exists, not just at point D.Now let us isothermally lower the pressure below point D, causing more liquid to vaporize. Eventually, wereach point F in figure , where the last drop of liquid vaporizes. Below F, we have only vapour. For any pointon the line between D and F liquid and vapour phases coexist in equilibrium.Q. Two liquids A and B have the same molecular weight and form an ideal solution. The solution has a vapour pressure of 700 Torrs at 80ºC. It is distilled till 2/3rd of the solution (2/3rd moles out of total moles) is collected as condensate. The composition of the condensate is xA = 0.75 and that of the residue is xA= 0.30. If the vapour pressure of the residue at 80ºC is 600 Torrs, find the original composition of the liquid ?

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The vapour pressure of water at 20 °C is 17.5 torr. What will be the no. of moles of water present in one litre of air at 20 °C and 40% relative humidity.(R =0.082 L atm/mol-k)a)4.2 x 10-4moleb)4.2 x 10-6molec)3.82 X 10-4 moled)3.82 X 10-6moleCorrect answer is option 'C'. Can you explain this answer?
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The vapour pressure of water at 20 °C is 17.5 torr. What will be the no. of moles of water present in one litre of air at 20 °C and 40% relative humidity.(R =0.082 L atm/mol-k)a)4.2 x 10-4moleb)4.2 x 10-6molec)3.82 X 10-4 moled)3.82 X 10-6moleCorrect answer is option 'C'. Can you explain this answer? for JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about The vapour pressure of water at 20 °C is 17.5 torr. What will be the no. of moles of water present in one litre of air at 20 °C and 40% relative humidity.(R =0.082 L atm/mol-k)a)4.2 x 10-4moleb)4.2 x 10-6molec)3.82 X 10-4 moled)3.82 X 10-6moleCorrect answer is option 'C'. Can you explain this answer? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for The vapour pressure of water at 20 °C is 17.5 torr. What will be the no. of moles of water present in one litre of air at 20 °C and 40% relative humidity.(R =0.082 L atm/mol-k)a)4.2 x 10-4moleb)4.2 x 10-6molec)3.82 X 10-4 moled)3.82 X 10-6moleCorrect answer is option 'C'. Can you explain this answer?.
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